Volume 27, No. 2, May 2014
http://dx.doi.org/10.14403/jcms.2014.27.2.297
SOME PROPERTIES OF DERIVATIONS ON CI-ALGEBRAS
Yong Hoon Lee* and Min Surp Rhee**
Abstract. The present paper gives the notion of a derivation on a CI-algebra X and investigates related properties. We define a set F ixd(X) by F ixd(X) = {x ∈ X | d(x) = x}, where d is a derivation on a CI-algebra X. We show that F ixd(X) is a subalgebra of X. Also, we prove some one-to-one and onto derivation theorems.
Moreover, we study a regular derivation on a CI-algebra and an isotone derivation on a transitive CI-algebra.
1. Introduction
The notion of a BCK-algebra was proposed by Y. Imai and K. Is´eki in 1966 [1]. In the same year, K. Is´eki introduced the notion of a BCI- algebra [2], which is a generalization of a BCK-algebra. H. S. Kim and Y. H. Kim defined a BE-algebra as a dualization of generalization of a BCK-algebra [4]. In [5], B. L. Meng introduced the notion of a CI- algebra as a generalization of a BE-algebra. Y. B. Jun and X. L. Xin applied the notion of a derivation on a BCI-algebra which is defined in a way similar to the notion of derivation in ring and near-ring theory [3].
In fact, the notion of a derivation in ring theory is quite old and plays a significant role in analysis, algebraic geometry and algebra. In this paper we introduce the notion of a derivation on a CI-algebra and obtain some properties related to it. In Section 2, we recall some definitions and some properties for a CI-algebra. In Section 3, we not only prove some one-to- one and onto derivation theorems, but also we study a regular derivation on a CI-algebra and an isotone derivation on a transitive CI-algebra.
Received March 03, 2014; Accepted April 07, 2014.
2010 Mathematics Subject Classification: Primary 06F35, 03G25, 08A30.
Key words and phrases: CI-algebra, self-distributive, derivation, transitive, iso- tone.
Correspondence should be addressed to Min Surp Rhee, [email protected].
2. Preliminaries
We review some definitions and properties that will be used later.
Definition 2.1. [5] Let X be a nonempty set, and let ∗ be a binary operation on X. Then (X, ∗, 1) is said to be a CI-algebra if the following axioms are satisfied:
(CI1) x ∗ x = 1.
(CI2) 1 ∗ x = x.
(CI3) x ∗ (y ∗ z) = y ∗ (x ∗ z) for all x, y, z ∈ X.
In the above definition we simply say that X is a CI-algebra. In [6], the author defined a binary relation ≤ by x ≤ y if and only if x ∗ y = 1, where x, y ∈ X.
A CI-algebra X is said to be self-distributive if x∗(y∗z) = (x∗y)∗(x∗z) for all x, y, z ∈ X [5]. A nonempty subset S of a CI-algebra X is said to be a subalgebra of X if x ∗ y ∈ S for all x, y ∈ S. Also, we define a binary operation ∨ on a CI-algebra X by x ∨ y = (y ∗ x) ∗ x.
Proposition 2.2. [5] Let x, y be arbitrary elements of a CI-algebra X. Then:
(1) y ∗ ((y ∗ x) ∗ x) = 1.
(2) (x ∗ 1) ∗ (y ∗ 1) = (x ∗ y) ∗ 1.
(3) 1 ≤ x implies x = 1 for all x, y ∈ X.
Definition 2.3. [5] A CI-algebra X is said to be transitive if (y ∗ z) ∗ ((x ∗ y) ∗ (x ∗ z)) = 1, for all x, y, z ∈ X.
Example 2.4. Let X = {1, a, b, c} be a CI-algebra, where ∗ is defined as follows:
∗ 1 a b c 1 1 a b c a 1 1 a a b 1 1 1 a c 1 1 a 1 Then X is a transitive CI-algebra.
Note that if X is a self-distributive CI-algebra, then X is a transitive CI-algebra.
Proposition 2.5. [7] Let X be a CI-algebra. If X is transitive, then:
(1) y ≤ z implies x ∗ y ≤ x ∗ z.
(2) y ≤ z implies z ∗ x ≤ y ∗ x for all x, y, z ∈ X.
Definition 2.6. [7] A nonempty subset of F of a CI-algebra X is said to be a filter of X if the following axioms are satisfied:
(F1) 1 ∈ F .
(F2) x ∈ F and x ∗ y ∈ F imply y ∈ F .
3. Derivations on CI-algebras
We start this section with the definition of a derivation on a CI- algebra X.
Definition 3.1. [3] Let X be a CI-algebra. A map d : X → X is said to be a right-left derivation ((r, l)-derivation) on X if it satisfies the identity
d(x ∗ y) = (x ∗ d(y)) ∨ (d(x) ∗ y)
for all x, y ∈ X. Similarly, a map d is said to be a left-right derivation ((l, r)-derivation) on X if it satisfies the identity
d(x ∗ y) = (d(x) ∗ y) ∨ (x ∗ d(y))
for all x, y ∈ X. Moreover, if d is both a (l, r)-derivation and a (r, l)- derivation, then d is called a derivation on X.
Example 3.2. Let X = {1, a, b} be a CI-algebra, where ∗ is defined as follows:
∗ 1 a b 1 1 a b a 1 1 b b 1 a 1 Then the map d : X → X defined by
d(x) = (
1 if x = 1, b a if x = a is a derivation on X.
Example 3.3. Let X = {1, a, b, c} be a CI-algebra, where ∗ is defined as follows:
∗ 1 a b c 1 1 a b c a 1 1 b a b 1 a 1 c c 1 1 b 1
Then the map d : X → X defined by
d(x) =
1 if x = 1, a a if x = c b if x = b is a derivation on X.
A map d on a CI-algebra X is said to be regular if d(1) = 1 [3]. Then we get two propositions for a regular map.
Proposition 3.4. Let d be a regular map on a CI-algebra X. Then the following properties are satisfied:
(1) If d is a (l, r)-derivation of X, then d(x) = x∨d(x) for every x ∈ X.
(2) If d is a (r, l)-derivation of X, then d(x) = d(x)∨x for every x ∈ X.
Proof. Note that d(1) = 1 and 1 ∗ x = x for every x ∈ X.
(1) If d is a (l, r)-derivation on X, then by Definition 3.1, we get d(x) = d(1 ∗ x) = (d(1) ∗ x) ∨ (1 ∗ d(x)) = x ∨ d(x)
for every x ∈ X.
(2) If d is a (r, l)-derivation on X, then by Definition 3.1, we get d(x) = d(1 ∗ x) = (1 ∗ d(x)) ∨ (d(1) ∗ x) = d(x) ∨ x
for every x ∈ X.
Proposition 3.5. Let d be a derivation on a CI-algebra X. If x∗1 = 1 for every x ∈ X, then d(1) = 1. That is, d is a regular derivation on X.
Proof. Let d be a (r, l)-derivation on X. Then by hypothesis we have d(1) = d(x ∗ 1)
= (x ∗ d(1)) ∨ (d(x) ∗ 1) (since Definition 3.1)
= (x ∗ d(1)) ∨ 1
= (1 ∗ (x ∗ d(1))) ∗ (x ∗ d(1)) (since x ∨ y = (y ∗ x) ∗ x)
= (x ∗ d(1)) ∗ (x ∗ d(1)) = 1
for every x ∈ X. Hence d is a regular (r, l)-derivation on X.
Similarly, let d be a (l, r)-derivation on X. Then by hypothesis we have d(1) = d(x ∗ 1)
= (d(x) ∗ 1) ∨ (x ∗ d(1)) (since Definition 3.1)
= 1 ∨ (x ∗ d(1)) (since x ∨ y = (y ∗ x) ∗ x)
= 1
for every x ∈ X. Hence d is a regular (l, r)-derivation on X. Therefore d is a regular derivation on X.
Now, we study some properties of a derivation on a CI-algebra X.
Proposition 3.6. Let d be a derivation on a CI-algebra X and let x ∈ X. If x ∗ 1 = 1, then x ∗ ((d(x) ∗ x) ∗ x) = x ∗ ((x ∗ d(x)) ∗ d(x)).
Proof. By Proposition 3.4, d(x) = d(x)∨x = x∨d(x). Hence x∗d(x) = x ∗ (d(x) ∨ x) = x ∗ ((x ∗ d(x)) ∗ d(x)) and x ∗ d(x) = x ∗ (x ∨ d(x)) = x∗((d(x)∗x)∗x). Thus we get x∗((d(x)∗x)∗x) = x∗((x∗d(x))∗d(x)).
Proposition 3.7. Let d be a (r, l)-derivation on a CI-algebra X.
Then:
(1) x ≤ d(x) for every x ∈ X.
(2) If X is transitive, d(x) ∗ y ≤ x ∗ y ≤ x ∗ d(y) for all x, y ∈ X.
Proof. (1) By Definition 2.1 (CI3) and Proposition 3.4 (2), we have x ∗ d(x) = x ∗ (d(x) ∨ x) = x ∗ ((x ∗ d(x)) ∗ d(x))
= (x ∗ d(x)) ∗ (x ∗ d(x)) = 1 for every x ∈ X. Hence x ≤ d(x).
(2) By (1) and Proposition 2.5, we have d(x) ∗ y ≤ x ∗ y ≤ x ∗ d(y) for all x, y ∈ X.
Theorem 3.8. Let d be a (r, l)-derivation on a transitive CI-algebra X. Then d(x ∗ y) = x ∗ d(y) for all x, y ∈ X.
Proof. Let d be a (r, l)-derivation on X and let x, y ∈ X. Then by Proposition 3.7 (2) we have d(x) ∗ y ≤ x ∗ d(y). Hence we get
d(x ∗ y) = (x ∗ d(y)) ∨ (d(x) ∗ y)
= ((d(x) ∗ y) ∗ (x ∗ d(y))) ∗ (x ∗ d(y))
= 1 ∗ (x ∗ d(y)) = x ∗ d(y).
Proposition 3.9. Let d be a map on a CI-algebra X and let x, y ∈ X.
Then:
(1) If d is a (l, r)-derivation on X, then x ∗ d(y) ≤ d(x ∗ y).
(2) If d is a (r, l)-derivation on X, then d(x) ∗ y ≤ d(x ∗ y).
(3) If d is a regular (r, l)-derivation on X, then d(x ∗ d(x)) = 1.
Proof. (1) Let d be a (l, r)-derivation of X. Then note that d(x ∗ y) = (d(x) ∗ y) ∨ (x ∗ d(y))
= ((x ∗ d(y)) ∗ (d(x) ∗ y)) ∗ (d(x) ∗ y).
Hence (x∗d(y))∗d(x∗y) = (x∗d(y))∗(((x∗d(y))∗(d(x)∗y))∗(d(x)∗y)) = 1 for all x, y ∈ X and so we get x ∗ d(y) ≤ d(x ∗ y).
(2) Let d be a (r, l)-derivation on X. Then note that d(x ∗ y) = (x ∗ d(y)) ∨ (d(x) ∗ y)
= ((d(x) ∗ y) ∗ (x ∗ d(y))) ∗ (x ∗ d(y)).
Hence (d(x)∗y)∗d(x∗y) = (d(x)∗y)∗(((d(x)∗y)∗(x∗d(y)))∗(x∗d(y))) = 1 for all x, y ∈ X and so we get d(x) ∗ y ≤ d(x ∗ y).
(3) If d is a (r, l)-derivation on X, then by Proposition 3.7 (1) and by hypothesis, we get d(x ∗ d(x)) = d(1) = 1 for every x ∈ X.
Proposition 3.10. Let X be a CI-algebra. If d1, d2, · · · , dn are regular (r, l)-derivations on X, then x ≤ dn(dn−1(...(d2(d1(x)))...)) for every n ∈ N.
Proof. For n = 1, we get
d1(x) = d1(1∗x) = (1∗d1(x))∨(d1(1)∗x) = d1(x)∨x = (x∗d1(x))∗d1(x).
Then by Definition 2.1 we have
x ∗ d1(x) = x ∗ ((x ∗ d1(x)) ∗ d1(x)) = (x ∗ d1(x)) ∗ (x ∗ d1(x)) = 1.
Hence x ≤ d1(x).
Suppose that x ≤ dn(dn−1(...(d2(d1(x)))...)) for any positive integer n ≥ 2. For simplicity, let
Dn= dn(dn−1(...(d2(d1(x)))...)).
Then
dn+1(Dn) = dn+1(1 ∗ Dn) = (1 ∗ dn+1(Dn)) ∨ (dn+1(1) ∗ Dn)
= dn+1(Dn) ∨ Dn= (Dn∗ dn+1(Dn)) ∗ dn+1(Dn).
Hence Dn∗ Dn+1 = 1, which implies Dn ≤ Dn+1. By assumption, we get x ≤ Dn≤ Dn+1. Therefore Proposition 3.10 holds.
Let d be a map on a CI-algebra X. Define a set F ixd(X) by F ixd(X) = {x ∈ X | d(x) = x}.
Now, we write d2(x) for (d ◦ d)(x), where ◦ is a composition.
Proposition 3.11. Let d be a derivation on a CI-algebra X and let x, y ∈ F ixd(X). Then:
(1) d(x) ∈ F ixd(X). Hence F ixd(X) ⊂ F ixd2(X).
(2) x ∗ y ∈ F ixd(X). That is, F ixd(X) is a subalgebra of X.
(3) x ∨ y ∈ F ixd(X).
Proof. (1) Note that d2(x) = d(d(x)) = d(x) = x for every x ∈ F ixd(X). Hence d(x) ∈ F ixd(X) and x ∈ F ixd2(X). Thus F ixd(X) ⊂ F ixd2(X).
(2) Let x, y ∈ F ixd(X). Then we get d(x) = x and d(y) = y. Hence d(x ∗ y) = (x ∗ d(y)) ∨ (d(x) ∗ y) = (x ∗ y) ∨ (x ∗ y) = x ∗ y.
So x ∗ y ∈ F ixd(X). Therefore F ixd(X) is a subalgebra of X.
(3) Let x, y ∈ F ixd(X). Then we get d(x ∨ y) = d((y ∗ x) ∗ x)
= ((y ∗ x) ∗ d(x)) ∨ (d(y ∗ x) ∗ y)
= ((y ∗ x) ∗ x) ∨ ((y ∗ d(x) ∨ d(y) ∗ x) ∗ x)
= ((y ∗ x) ∗ x) ∨ ((y ∗ x) ∗ x)
= (y ∗ x) ∗ x = x ∨ y since x ∨ x = (x ∗ x) ∗ x = 1 ∗ x = x.
In the above proposition we say that d is idempotent if d2(x) = x.
Proposition 3.12. Let d be a (r, l)-derivation of a self-distributive CI-algebra X. If y ∈ F ixd(X), then x ∗ y ∈ F ixd(X) for every x ∈ X.
Proof. Let y ∈ F ixd(X). Then d(y) = y. By Definition 2.1 (CI3) and Proposition 3.7 (1), we have
d(x ∗ y) = (x ∗ d(y)) ∨ (d(x) ∗ y) = ((d(x) ∗ y) ∗ (x ∗ y)) ∗ (x ∗ y)
= (x ∗ ((d(x) ∗ y) ∗ y)) ∗ (x ∗ y)
= ((x ∗ (d(x) ∗ y)) ∗ (x ∗ y)) ∗ (x ∗ y)
= (((x ∗ d(x)) ∗ (x ∗ y)) ∗ (x ∗ y)) ∗ (x ∗ y)
= ((1 ∗ (x ∗ y)) ∗ (x ∗ y)) ∗ (x ∗ y)
= ((x ∗ y) ∗ (x ∗ y)) ∗ (x ∗ y) = 1 ∗ (x ∗ y) = x ∗ y.
This completes the proof.
By Proposition 3.11 (1) we know F ixd(X) is a subset of F ixd2(X).
But in general F ixd2(X) is not a subset of F ixd(X). Also, if x ∈ F ixd(X), then d(x) ∈ F ixd(X). But there is an element x /∈ F ixd(X) even though d(x) ∈ F ixd(X). In fact, d(b) ∈ F ixd(X) but b /∈ F ixd(X) in Example 3.2.
Theorem 3.13. Let d be a derivation on a CI-algebra X. If d(x) ∈ F ixd(X) for every x ∈ X, then the following conditions are equivalent:
(1) d(x) = x for every x ∈ X.
(2) d is a one-to-one derivation.
(3) d is a onto derivation.
Proof. (1) ⇒(2) and (1) ⇒(3) are clear.
(2) ⇒(1) Let d be a one-to-one derivation and let x ∈ X. Since d(x) ∈ F ixd(X), we have d(d(x)) = d(x). Since d is one-to-one, we get d(x) = x for every x ∈ X.
(3) ⇒(1) Let d be a onto derivation and let x ∈ X. Then there exists y ∈ X such that d(y) = x. Since d(x) ∈ F ixd(X), we have d(x) = d(d(y)) = d(y) = x.
Let d be a map on a CI-algebra X. Define the kernel of d, denoted by Kerd,
Kerd = {x ∈ X | d(x) = 1}.
Proposition 3.14. Let d be a (r, l)-derivation on a CI-algebra X. If x ∗ 1 = 1 for every x ∈ X, then Kerd is a subalgebra of X.
Proof. Let x, y ∈ Kerd. Then d(x) = 1 and d(y) = 1. Hence we have d(x ∗ y) = (x ∗ d(y)) ∨ (d(x) ∗ y) = (x ∗ 1) ∨ (1 ∗ y) = 1 ∨ y = 1.
So x ∗ y ∈ Kerd. Therefore Kerd is a subalgebra of X.
A CI-algebra X is said to be commutative if (y ∗ x) ∗ x = (x ∗ y) ∗ y for all x, y ∈ X.
Proposition 3.15. Let d be a (r, l)-derivation on a commutative CI- algebra X and let x ≤ y. If x ∈ Kerd and z ∗ 1 = 1 for every z ∈ X, then y ∈ Kerd.
Proof. Let x ∈ Kerd and let x ≤ y. Then d(x) = 1 and x ∗ y = 1. By hypothesis and Proposition 2.2 (2), we get
d(y) = d(1 ∗ y) = d((x ∗ y) ∗ y) = d((y ∗ x) ∗ x)
= ((y ∗ x) ∗ d(x)) ∨ (d(y ∗ x) ∗ x)
= ((y ∗ x) ∗ 1) ∨ (d(y ∗ x) ∗ x)
= 1 ∨ (d(y ∗ x) ∗ x) = 1, and so y ∈ Kerd.
A map d on a CI-algebra X is said to be an isotone if x ≤ y implies d(x) ≤ d(y) for all x, y ∈ X [8].
Theorem 3.16. Let d be an isotone derivation on a transitive CI- algebra X and let x ∈ F ixd(X). If d is idempotent and regular on X, then Kerd is a filter of X.
Proof. Since d is regular, we get 1 ∈ Kerd. Let x ∈ Kerd and x ∗ y ∈ Kerd. Then by Theorem 3.8, we get 1 = d(x ∗ y) = x ∗ d(y), which means x ≤ d(y). Since d is isotone and idempotent derivation on X, then 1 = d(x) ≤ d2(y) = d(y). That is, d(y) = 1. This implies y ∈ Kerd. Therefore Kerd is a filter of X.
Proposition 3.17. Let d be a regular derivation on a CI-algebra X.
If d is one-to-one, then Kerd = {1}.
Proof. If x ∈ Kerd, then d(x) = 1. Since d is regular, d(1) = 1.
Hence d(x) = d(1). Since d is one-to-one, we get x = 1. Therefore Kerd = {1}.
Proposition 3.18. Let d be an isotone derivation on a CI-algebra X. If x ≤ y and x ∈ Kerd, then y ∈ Kerd for all x, y ∈ X.
Proof. Let x ≤ y and x ∈ Kerd. Then d(x) = 1, and so 1 = d(x) ≤ d(y). By Proposition 2.2 (3), we get d(y) = 1.
Definition 3.19. [3] Let d be a derivation on a CI-algebra X. A nonempty subset F of X is said to be d-invariant if d(F ) ⊆ F , where d(F ) = {d(x) | x ∈ F }.
It follows from d(F ixd(X)) ⊂ F ixd(X) that F ixd(X) is d-invariant.
Proposition 3.20. Let d be a regular derivation on a CI-algebra X.
Then every filter F of X is d-invariant.
Proof. Let F be a filter of X and let y ∈ d(F ). Then y = d(x) for some x ∈ F. It follows from Proposition 3.9 that x∗y = x∗d(x) = 1 ∈ F.
This implies y ∈ F. Thus d(F ) ⊆ F. Therefore F is d-invariant.
Definition 3.21. [7] A nonempty subset F of a CI-algebra X is said to be a normal filter of X if the following conditions are satisfied:
(NF1) 1 ∈ F.
(NF2) x ∈ X and y ∈ F imply x ∗ y ∈ F.
Proposition 3.22. Let d be a (r, l)-derivation on a self-distributive CI-algebra X. Then:
(1) F ixd(X) is a normal filter of X.
(2) Kerd is a normal filter of X.
Proof. Since X is a self-distributive CI-algebra, then x ∗ 1 = x ∗ (x ∗ x) = (x ∗ x) ∗ (x ∗ x) = 1 for every x ∈ X. By Proposition 3.5 we know d(1) = 1. Hence 1 ∈ F ixd(X) and so 1 ∈ Kerd.
(1) Let x ∈ X and y ∈ F ixd(X). Then by Theorem 3.8, we get d(x ∗ y) = x ∗ d(y) = x ∗ y. Hence x ∗ y ∈ F ixd(X) and so F ixd(X) is a normal filter of X.
(2) Similarly, let x ∈ X and y ∈ Kerd. Then by Theorem 3.8, we get d(x ∗ y) = x ∗ d(y) = x ∗ 1 = 1. Hence x ∗ y ∈ Kerd. and so Kerd is a normal filter of X.
Proposition 3.23. Let d be a regular derivation on a self-distributive CI-algebra X and let a ∈ X. Then Fa= {x ∈ X | a ≤ d(x)} is a normal filter of X.
Proof. Since a ≤ d(1) = 1 for every a ∈ X, we get 1 ∈ Fa. Let x ∈ X and y ∈ Fa. Then a ∗ d(y) = 1, and so by Theorem 3.8
a ∗ d(x ∗ y) = a ∗ (x ∗ d(y)) = (a ∗ x) ∗ (a ∗ d(y))
= (a ∗ x) ∗ 1
= 1.
Hence x ∗ y ∈ Fa and so Fa is a normal filter of X.
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*
Department of Mathematics Dankook University
Cheonan 330-714, Republic of Korea E-mail: [email protected]
**
Department of Mathematics Dankook University
Cheonan 330-714, Republic of Korea E-mail: [email protected]
